JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
Let \(A\) be a \(3\times3\) matrix such that \(A\left[ {\begin{array}{*{20}{c}}
1&2&3 \\
0&2&3 \\
0&1&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0&1 \\
1&0&0 \\
0&1&0
\end{array}} \right]\) Then \(A^{-1}\) is
- A \(\left[ {\begin{array}{*{20}{c}}
3&1&2 \\
3&0&2 \\
1&0&1
\end{array}} \right]\) - B \(\left[ {\begin{array}{*{20}{c}}
3&2&1 \\
3&2&0 \\
1&1&0
\end{array}} \right]\) - C \(\left[ {\begin{array}{*{20}{c}}
0&1&3 \\
0&2&3 \\
1&1&1
\end{array}} \right]\) - D \(\left[ {\begin{array}{*{20}{c}}
1&2&3 \\
0&1&1 \\
0&2&3
\end{array}} \right]\)
Answer & Solution
Correct Answer
(A) \(\left[ {\begin{array}{*{20}{c}}
3&1&2 \\
3&0&2 \\
1&0&1
\end{array}} \right]\)
Step-by-step Solution
Detailed explanation
Given \(A\left[ {\begin{array}{*{20}{c}} 1&2&3\\ 0&2&3\\ 0&1&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0&1\\ 1&0&0\\ 0&1&0 \end{array}} \right]\) Applying \({C_1} \leftrightarrow {C_3}\)…
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